# Thread: integrals #2

1. ## integrals #2

Please help me with these problems. Thank you.
1. integral from 0 to pi of (4 sin theta -3 cos theta) d theta
I started like this...4 integral from 0 to pi of sin theta d theta -3 integral from 0 to pi cos theta d theta...I got stuck here
2. How do I calculate this math?
3/4 (8^4/3) - 3/4 (1^4/3) =?
3. How do I solve this integral?
integral of cos^3 x dx
Thanks again for your help!

2. 1. integral from 0 to pi of (4 sin theta -3 cos theta) d theta
I started like this...4 integral from 0 to pi of sin theta d theta -3 integral from 0 to pi cos theta d theta...I got stuck here

what do you mean you "got stuck"? do you know the antiderivatives of sine and cosine ? do you know how to use the fundamental theorem of calculus?

2. How do I calculate this math?
3/4 (8^4/3) - 3/4 (1^4/3) =?

$8^{\frac{4}{3}} = \left(8^{\frac{1}{3}}\right)^4 = 2^4 = 16$

1 to any power is 1.

3. How do I solve this integral?
integral of cos^3 x dx

$cos^3{x} = cos^2{x} \cos{x} = (1 - \sin^2{x})\cos{x} = \cos{x} - \sin^2{x}\cos{x}$

$\int \cos{x} \, dx - \int \sin^2{x}\cos{x} \, dx$

straight up integration of the first integral ... use substitution for the second , let $u = \sin{x}$